of 1 31
Semiconductor Circuits, Stokes Theorem, And The Maxwell Equations
By
Ian Beardsley
Copyright © 2021 by Ian Beardsley
of 2 31
Table of Contents
Abstract……………………………………….3
Important……………………………………..4
The Computation…………………………….5
The Dynamic Function………………………5
Life……………………………………………9
The Delta-Phi Function…………………….10
The Path Integral……………………………15
The Constant…………………………………20
Summary……………………………………..22
Approximation……………………………….23
Appendix 1…………………………………..27
Appendix 2…………………………………..29
of 3 31
Abstract
It is suggested that artificial intelligence and biological life elements are mathematical constructs that
describe one another. This leads to a connection of the artificial intelligence elements to Stoke’s theorem
and Maxwell’s Equations.
of 4 31
Important
Above we see the artificial intelligence (AI) elements pulled out of the periodic table of the elements. As
you see we can make a 3 by 3 matrix of them and an AI periodic table. Silicon and germanium are in
group 14 meaning they have 4 valence electrons and want 4 for more to attain noble gas electron
configuration. If we dope Si with B from group 13 it gets three of the four electrons and thus has a
deficiency becoming positive type silicon and thus conducts. If we dope the Si with P from group 15 it
has an extra electron and thus conducts as well. If we join the two types of silicon we have a
semiconductor for making diodes and transistors from which we can make logic circuits for AI.
As you can see doping agents As and Ga are on either side of Ge, and doping agent P is to the right of Si
but doping agent B is not directly to the left, aluminum Al is. This becomes important. I call (As-Ga) the
differential across Ge, and (P-Al) the differential across Si and call Al a dummy in the differential because
boron B is actually used to make positive type silicon.
That the AI elements make a three by three matrix they can be organized with the letter E with subscripts
that tell what element it is and it properties, I have done this:
Thus E24 is in the second row and has 4 valence electrons making it silicon (Si), E14 is in the first row
and has 4 valence electrons making it carbon (C). I believe that the AI elements can be organized in a 3 by
3 matrix makes them pivotal to structure in the Universe because we live in three dimensional space so
the mechanics of the realm we experience are described by such a matrix, for example the cross product.
Hence this paper where I show AI and biological life are mathematical constructs and described in terms
of one another.
We see, if we include the two biological elements in the matrix (E14) and and (E15) which are carbon and
nitrogen respectively, there is every reason to proceed with this paper if the idea is to show not only are
the AI elements and biological elements mathematical constructs, they are described in terms of one
another. We see this because the first row is ( B, C, N) and these happen to be the only elements that are
not core AI elements in the matrix, except boron (B) which is out of place, and aluminum (Al) as we will
see if a dummy representative, makes for a mathematical construct, the harmonic mean. Which means we
have proved our case because the first row if we take the cross product between the second and third rows
are, its respective unit vectors for the components, meaning they describe them!
E
13
E
14
E
15
E
23
E
24
E
25
E
33
E
34
E
35
of 5 31
The Computation
And silicon (Si) is at the center of our AI periodic table of the elements. We see the biological elements C
and N being the unit vectors are multiplied by the AI elements, meaning they describe them! But we have
to ask; Why does the first row have boron in it which is not a core biological element, but is a core AI
element? The answer is that boron is the one AI element that is out of place, that is, aluminum is in its
place. But we see this has a dynamic function.
The Dynamic Function
The primary elements of artificial intelligence (AI) used to make diodes and transistors, silicon (Si) and
germanium (Ge) doped with boron (B) and phosphorus (P) or gallium (Ga) and arsenic (As) have an
asymmetry due to boron. Silicon and germanium are in group 14 like carbon (C) and as such have 4
valence electrons. Thus to have positive type silicon and germanium, they need doping agents from group
13 (three valence electrons) like boron and gallium, and to have negative type silicon and germanium they
need doping agents from group 15 like phosphorus and arsenic. But where gallium and arsenic are in the
same period as germanium, boron is in a different period than silicon (period 2) while phosphorus is not
(period 3). Thus aluminum (Al) is in boron’s place. This results in an interesting equation.
A = (Al, Si, P )
B = (G a, G e, As)
A ×
B =
B
C
N
Al Si P
G a Ge As
= (Si ⋅ As − P ⋅ G e)
B + (P ⋅ G a − Al ⋅ As)
C + (Al ⋅ G e − Si ⋅ G a)
N
A = 26.98
2
+ 28.09
2
+ 30.97
2
= 50g /m ol
B = 69.72
2
+ 72.64
2
+ 74.92
2
= 126g /m ol
A ⋅
B = A Bcosθ
cosθ =
6241
6300
= 0.99
θ = 8
∘
A ×
B = A Bsi nθ = (50)(126)sin8
∘
= 877.79
877.79 = 29.6g /m ol ≈ Si = 28.09g /m ol
Si(A s − G a) + G e(P − Al )
SiG e
=
2B
Ge + Si
of 6 31
The differential across germanium crossed with silicon plus the differential across silicon crossed with
germanium normalized by the product between silicon and germanium is equal to the boron divided by
the average between the germanium and the silicon. The equation has nearly 100% accuracy (note: using
an older value for Ge here, is now 72.64 but that make the equation have a higher accuracy):
We found (Beardsley, Mathematical Structure, 2020) that the differential across silicon (P-Al) times
germanium (Ge) over boron (B) plus the differential across germanium (As-Ga) times silicon (Si) over
boron (B) was equal to the harmonic mean between Si and Ge. This was interesting because aluminum is
used as what I called a dummy doping agent element, which when inserted predicts the actually doping
agent boron, that seems out of place in the periodic table where the core artificial intelligence elements are
concerned. This is written:
Thus because boron is out of place we get the harmonic mean between core AI elements Si and Ge on the
right. But one the left we have a difference between doping elements times a ratio plus another difference
between doping agent times a ratio. The ratios are the semiconducting elements. But the one associated
with Ge is multiplied by Si and the one associated with Si is multiplied by Ge. We have seen this pattern
before, it is stokes theorem:
Stokes Theorem states:
Where:
Thus we have…
28.09(74.92 − 69.72) + 72.61(30.97 − 26.98)
(28.09)(72.61)
=
2(10.81)
(72.61 + 28.09)
0.213658912 = 0.21469712
0.213658912
0.21469712
= 0.995
Si
B
(As − G a) +
Ge
B
(P − Al ) =
2SiG e
Si + G e
∫
S
( ∇ × u ) ⋅ d S =
∮
C
u ⋅ d r
∇ × u =
i
j k
∂
∂x
∂
∂ y
∂
∂z
u
1
u
2
u
3
of 7 31
We know the harmonic mean H of a function is
And, that the arithmetic mean A of a function is
We have
But, we want to use Stokes theorem so we want the integral in the numerator. So, we make the
approximation
And, we have
But, this is only 80% accurate. We find it is very accurate if we say
i
j
k
∂
∂x
∂
∂ y
∂
∂z
0
Si
B
(Ga)z
Si
B
(As)y
=
Si
B
(As − G a)
i
i
j
k
∂
∂x
∂
∂ y
∂
∂z
Ge
B
(Al )z 0
Ge
B
(P)x
=
Ge
B
(P − Al )
j
H =
1
1
b − a
∫
b
a
f (x)
−1
d x
A =
1
b − a
∫
b
a
f (x)d x
Si
B
(As − G a) +
Ge
B
(P − Al ) =
Ge − Si
Ge
Si
dx
x
H ≈ A
∫
1
0
∫
1
0
[
Si
B
(As − G a) +
Ge
B
(P − Al )
]
d xd y ≈
1
Ge − Si
∫
Ge
Si
x d x
of 8 31
Which yields
We have by molar mass
Thus,…
We can break up our integral into two integrals u, and v:
Where 1/3 is approximately and 2/3 is approximately and it has the accuracy of the
80% version where we made H approximately A.
f (x) =
4
5
x
∫
1
0
∫
1
0
[
Si
B
(As − G a) +
Ge
B
(P − Al )
]
d xd y ≈
1
Ge − Si
4
5
∫
Ge
Si
x d x
Si
B
(Ga) =
28.09
10.81
(69.72) = 181.1688g/m ol
Ge
B
(Al ) =
72.61
10,81
(26.98) = 181.2227g/m ol
Si
B
(As) =
28.09
10.81
(74.92) = 194.68111g/m ol
Ge
B
=
72.61
10.81
(30.97) = 208.02328g/m ol
u = 181z
j + 195y
k
v = 181z
i + 208y
k
∫
1
0
∫
1
0
Si
B
(As − G a)d yd z ≈
1
3
1
(Ge − Si )
∫
Ge
Si
x d x
∫
1
0
∫
1
0
Ge
B
(P − Al ) d x dz ≈
2
3
1
(Ge − Si )
∫
Ge
Si
yd y
(1 − ϕ)
ϕ
of 9 31
Life
In order to have life you need carbon because it has four valence electrons allowing it to form in
long chains with hydrogen, nitrogen and oxygen. Silicon is in the same group as carbon and
therefore has 4 valence electrons as well. However it cannot form long chains with hydrogen
because in the presence of other elements it reacts with them, like with O2 to make SiO2 or sand.
Silicon has been considered in its possibility to make life along with boron polymers (Mann and
Perry 1986; Trevors 1997a; Williams 1986) with the conclusion that both silicon and boron lack
”replicative potential”.
Conclusion
It would seem boron is on the dividing line between silicon based life as electronic based life,
and biological life. This puts it next to carbon in the periodic table and diagonal to silicon,
aluminum it its place. If we use aluminum as a dummy in the silicon differential, we have a
mathematical function that is the harmonic mean between silicon and germanium. This
mathematical dynamic seems to be integral in the mathematical relationship between AI and
biological life.
Note, I used an older version of the molar mass for Germanium, it has since been more
accurately determined to be 72.64 as opposed to 72.61, which only make the equation more
accurate.
I have also shown that Stokes form of the equation works for the elements in terms of density,
and atomic radius which are not the harmonic mean but geometric and arithmetic means
respectively so the equation takes a generalized form of:
The power mean is obtained by letting
It is the geometric mean if
and by molar mass or
and by density or
and by atomic radius
Are respectively and
∇ ⋅ Q = C f
−1
(
1
n
⋅
n
∑
i=1
f (x
i
)
)
f (x) = x
p
f (x) = log(x)
(As − G a)
(P − Al )
(Al − P)
(Al − P)
(Al − P)
(Ga − As)
ΔE
1
ΔE
2
of 10 31
And, the ratios
and by molar mass or
and by density or
and by atomic radius where C is
Are, quotients , and , respectively, then
The Delta-Phi Function
If we take our equation
And make the approximation:
Which is:
Is close to 91% accurate
And write (As-Ga) is the differential across Ge as and (P-Al) is the differential across Si and write it
, then we have:
Then we have
Si
B
Ge
B
B
2Ge(Ga − As)
Si
B
2Ge(Ga − As)
P
Si
B
Ge
B
Φ
Q
1
Q
2
∇ = (ΔE
1
, ΔE
2
)
Q = (Q
1
, Q
2
)
Si(A s − G a)
B
+
Ge(P − Al )
B
=
2SiG e
Si + G e
2SiG e
Si + G e
≈ Ge − Si
2SiG e
Si + G e
=
2(28.09)(72.64)
28.09 + 72.64
= 40.5g /m ol
Ge − Si = 72.64 − 28.09 = 44.55g /m ol
40.5
44.55
100 = 90.9
ΔGe
ΔSi
(Si )ΔG e + (G e)ΔSi = (G e − Si )B
of 11 31
Which has an accuracy of
and
B=10.81 g/mol
9.7797/10.81(100)-=90%
But here we notice than 0.63 is approximately the golden ratio conjugate (phi), and 1.63 is
approximately the golden ratio (Phi) where
and
And
=1.618
=0.618
So we have the equation
With an accuracy of 89.4%
I call this the Delta-Phi function, ( is delta).
If we denote not the horizontal changes across Si and Ge in the periodic table which are differences
between doping agents we outlined before, but the vertical change in the periodic table from Ge to Si
which are the semiconductor materials themselves as , then we have:
I am thinking this is actually
Si
(Ge − Si )
ΔGe +
Ge
(Ge − Si )
ΔSi = B
0.63ΔGe + 1.63ΔSi = B
ΔGe = 5.2
ΔSi = 3.99
ϕ
Φ
a = b + c
a
b
=
b
c
Φ = a /b
ϕ = b /a
(ϕ)ΔGe + (Φ)ΔSi = B
Δ
ΔS
Si
ΔGe
ΔS
+ G e
ΔSi
ΔS
= B
Si
dGe
dS
+ Ge
dSi
dS
= B
of 12 31
The derivatives of some functions evaluated at some important points. One then has to ask what role does
molar mass play in semiconduction, One might ask what are the roles played by density and atomic radius
since the equation generalizes to these with the f-mean.
Which is quite interesting because it says the nature of the strange placement of boron is actually the
change across Ge times Si plus the the change across Si times Ge, all that divided by the difference
between Ge and Si, We can see its dynamics in the following illustration:
If we have
And that
And actually this is:
Where
Si
ΔGe
ΔS
+ G e
ΔSi
ΔS
= B
ΔGe
ΔS
=
5.2
44.5
= 0.116865
ΔSi
ΔS
=
3.99
44.5
= 0.8966
Si
df
dx
+ G e
dg
dx
= B
of 13 31
Then f prime of x and g prime of x can perhaps be found by by considering the the elements above and
below the elements in what I am calling the AI periodic table of the elements. We apply the same process
of comparing the change in the horizontal differentials down through periods 2 to 6 in the periodic table
of the elements, and the semiconductor differences down through the group 14 elements in the periodic
table of the elements. And plotting them. We find, if we consider the trend to decrease and not jut up at
x=4 for g prime of x and and x=3 for x prime of x, then our functions are exponentially decreasing by
some factor of some function for each, as you can see in the data tables and graphs in the next couple of
pages. The functions are:
If we have
But something very interesting is happening here. The Maxwell-Faraday Equation is in relating the
electric field to the magnetic field
Apply Stoke’s Theorem on the left
The integrands on both sides have to be equal at every point in space because the surfaces S over which
they are integrated are the same. Therefore
This differential form relates the change in the electric field with respect to position to the change in the
magnetic field with respect to time. But even more interesting is
f′ (0) = 0.11685
g′ (0) = 0.08966
g′ (x) =
2
3
e
−x
f′ (x) =
6
7
e
−x
g(x) = −
2
3
e
−x
f (x ) = −
6
7
e
−x
f′ (2) = 0.11685
g′ (2) = 0.08966
E
B
∮
C
E ⋅ d
l = −
∂
∂t
∫
S
B ⋅ d s
∫
S
( ∇ ×
E ) ⋅ d s = −
∂
∂t
∫
S
B ⋅ d s
∇ ×
E = −
∂
∂t
B
of 14 31
Ampere’s Circuit Law with Maxwell’s addition:
Remember we said
We can write this
We see that is like and is like meaning that boron B is like .
Where is current density and boron B the interesting out of place element in the periodic table. This is
interesting because we are not dealing with E and B fields, but molar masses of semiconductor elements
and their doping agents. Essentially it says just as the change in the magnetic field is related to the change
in the electric field, so is the the change in doping agents with respect to semiconductor materials of
silicon to those changes in those that are semiconductor materials of germanium. In a sense the
relationship between E and B fields which are just different manifestations of one another are as are the
relationships between silicon and and germanium that are different manifestations of one another.
∇ ×
B = μ
0
J + μ
0
ϵ
∂
E
∂t
Si
ΔGe
ΔS
+ G e
ΔSi
ΔS
= B
Si
ΔGe
ΔS
= B − G e
ΔSi
ΔS
Si
ΔGe
ΔS
∇ ×
B
Ge
ΔSi
ΔS
μ
0
ϵ
∂
E
∂t
μ
0
J
J
of 15 31
The Path Integral
We have
Which is
Let’s find the geometric interpretation and find the path over which the integral is done. Thus
we have
, ,
∫
1
0
∫
1
0
Si
B
(As − G a)d yd z ≈
1
3
1
(Ge − Si )
∫
Ge
Si
x d x
∫
1
0
∫
1
0
Ge
B
(P − Al ) d x dz ≈
2
3
1
(Ge − Si )
∫
Ge
Si
yd y
∫
S
( ∇ × u ) ⋅ d S =
∮
C
u ⋅ d r
r
i
j
k
∂
∂x
∂
∂ y
∂
∂z
0
Si
B
(Ga)z
Si
B
(As)y
=
Si
B
(As − G a)
i
i
j
k
∂
∂x
∂
∂ y
∂
∂z
Ge
B
(Al )z 0
Ge
B
(P)x
=
Ge
B
(P − Al )
j
Si
B
(Ga) =
28.09
10.81
(69.72) = 181.1688g/m ol
Si
B
(As) =
28.09
10.81
(74.92) = 194.68111g/m ol
u
x
= 0
u
y
= 181z
u
z
= 195y
u = 181z
j + 195y
k
of 16 31
The geometric interpretation is as follows…
of 17 31
Let us change x to z on the right:
Then,..
,
Is a line in the y-z plane (See Appendix 2).
0.618 is exactly the the golden ratio conjugate. Thus the square root of C is precisely this value
reduced by a factor of 100. That is…
At this point it would be good to note that Si+Ge=100.73 g/mol, almost exactly 100.
u =
(
0,
Si
B
(Ga)z,
Si
B
(As)y
)
(
0,
Si
B
(Ga)z,
Si
B
(As)y
)
⋅ (d x, d y, dz) =
1
3
x d x
Ge − Si
d r = (d x, d y, d z)
(
0,
Si
B
(Ga)z,
Si
B
(As)y
)
⋅ (d x, d y, dz) =
1
3
zdz
Ge − Si
d x = 0
dz = 0
d y =
Bd z
3(Si )(Ga)(Ge − Si )
r = (t)
j + (26,178t)
k
y =
10.81
3(28.09)(69.72)(72.64 − 28.09)
z
y = Cz
C = 0.0000382
mol
2
g
2
C = 0.00618m ol /g
C =
ϕ
100
of 18 31
We can also write
,
In which case we have
In which case
We can also find two solutions for the other integral,…
In which case we have
(
0,
Si
B
(Ga)z,
Si
B
(As)y
)
⋅ (d x, d y, dz) =
1
3
yd y
Ge − Si
d x = 0
d y = 0
dz =
Bd y
3(Si )(As)(Ge − Si )
z =
(10.81)
3(28.09)(74.92)(72.64 − 28.09)
y
C = 0.0000384
mol
2
g
2
C = 0.0062m ol /g
z = C y
∫
1
0
∫
1
0
Ge
B
(P − Al ) d x dz ≈
2
3
1
(Ge − Si )
∫
Ge
Si
yd y
(
0,
Ge
B
(Al )z,
Ge
B
(P)y
)
⋅ (d x, d y, dz) =
2
3
zdz
Ge − Si
d y =
2Bd z
3(Si )(Al )(Ge − Si )
y =
2(10.81)
3(28.09)(26.98)(72.64 − 28.09)
z
C = 0.0002
mol
2
g
2
of 19 31
And the other solution is:
Thus we have the two integrals each with two solutions:
1.
2.
1.
2.
Since the line y=(0.0000382) and y=(1/0.0000382)z represent slope inverses, then they are 45
degrees apart. All in all this brings to mind the Stokes Parameters used in measuring polarized
light, because instead of having two lines 90 degrees apart they have 4 lines 45 degrees apart, the
, , , axes.
C = 0.0146m ol /g
dz =
2Bd y
3(Si )(P)(Ge − Si )
y =
2(10.81)
3(28.09)(30.97)(72.64 − 28.09)
z
C = 0.0001859
mol
2
g
2
C = 0.0136m ol /g
∫
1
0
∫
1
0
Si
B
(As − G a)d yd z ≈
1
3
1
(Ge − Si )
∫
Ge
Si
x d x
y = (0.0000382)z
z = (0.0000382)y
∫
1
0
∫
1
0
Ge
B
(P − Al ) d x dz ≈
2
3
1
(Ge − Si )
∫
Ge
Si
yd y
y = (0.0002)z
z = (0.0002)y
I
x
I
y
I
x′
I
y′
of 20 31
The Constant
Let us revisit our value for the constant C on page 17:
Since Si+Ge=100.73 g/mol, almost exactly 100. And, we need grams per mole in the
denominator…
Which means that
Or,…
B=9.85
9.85/10.81=91% accuracy
The result
The Si and Ge are the core semiconductor elements and are central to our AI periodic table
which we have said is a 3 by 3 matrix and that 3 by 3 matrices are core to the physics of the
Universe we experience 3D space:
d y =
Bd z
3(Si )(Ga)(Ge − Si )
C =
ϕ
100
C =
ϕ
Si + Ge
B
3(Si )(Ga)(Ge − Si )
=
ϕ
Si + Ge
B = 3ϕ
2
SiGa
(Ge − Si )
(Ge + Si)
2
B = 3(0.618)
2
(28.09)(69.72)
(72.64 − 28.09)
(72.64 + 28.09)
2
C =
ϕ
Si + Ge
of 21 31
And that through it we have the biological elements describing the AI elements:
And, that this results in our dynamic equations:
Where
In the first equation changing dx to dz.
E
13
E
14
E
15
E
23
E
24
E
25
E
33
E
34
E
35
A ×
B =
B
C
N
Al Si P
Ga G e As
= (Si ⋅ As − P ⋅ Ge)
B + (P ⋅ G a − Al ⋅ As)
C + (Al ⋅ Ge − Si ⋅ Ga)
N
∫
1
0
∫
1
0
Si
B
(As − G a)d yd z ≈
1 − ϕ
(Ge − Si )
∫
Si
Ge
x d x
∫
1
0
∫
1
0
Ge
B
(P − Al ) d x dz ≈
ϕ
(Ge − Si )
∫
Si
Ge
yd y
r = (t)
j + (26,178t)
k
C =
B
3(Si )(Ga)(Ge − Si )
of 22 31
Summary
Due to an asymmetry in the periodic table of the elements due to boron we have the harmonic
mean between the semiconductor elements (by molar mass):
This is Stokes Theorem:
Which can be generalized to include the AI elements by density and atomic radius:
By making the approximation
In
We have
Which is Ampere’s Circuit Law
We see if written
Which is interesting because it is semiconductor elements by molar mass, which are used to
make circuits.
Si
B
(As − G a) +
Ge
B
(P − Al ) =
2SiG e
Si + G e
∫
S
( ∇ × u ) ⋅ d S =
∮
C
u ⋅ d r
∇ ⋅ Q = C f
−1
(
1
n
⋅
n
∑
i=1
f (x
i
)
)
2SiGe
Si + Ge
≈ Ge − Si
Si(As − G a)
B
+
Ge(P − Al )
B
=
2SiGe
Si + Ge
Si
ΔGe
ΔS
+ G e
ΔSi
ΔS
= B
∇ ×
B = μ
0
J + μ
0
ϵ
∂
E
∂t
Si
ΔGe
ΔS
= B − Ge
ΔSi
ΔS
of 23 31
This results in an interesting constant C which is the golden ratio conjugate phi divided by the
semiconductor elements if we take the square root of it:
Where (Phi) is given by
and
And
=1.618
=0.618
(phi) the golden ratio conjugate. We also find
Approximation
If in the equation
We make the approximation
Then the Stokes form of the equation becomes
Which is very accurate.
C =
ϕ
Si + Ge
Φ
a = b + c
a
b
=
b
c
Φ = a /b
ϕ = b /a
ϕ
(ϕ)ΔGe + (Φ)ΔSi = B
Si
B
(As − G a) +
Ge
B
(P − Al ) =
Ge − Si
Ge
Si
dx
x
2SiGe
Si + Ge
≈ Ge − Si
∫
1
0
∫
1
0
[
Si
B
(As − G a) +
Ge
B
(P − Al )
]
d yd z =
∫
Ge
Si
d x
of 24 31
Thus we see for this approximation there are two integrals as well:
For which the respective paths are
And the respective curl fluxes are
Which are in grams per mole and are nitrogen (N) 14 g/mol and aluminum (Al) 27 g/mol.
electronics. Crystalline Aluminum Nitride has piezoelectric properties, meaning it can generate
an electric charge if mechanically stressed. This means it can be used for surface acoustic wave
sensors deposited on silicon wafers. It is used for RF filters in mobile phones and promises to
perhaps have uses as a superconductor at high temperatures and there is much research to make
aluminum gallium nitride alloy light emitting diodes to operate in the ultraviolet, along with
gallium nitride semiconductors. The former has been done down to 250 nm.
∫
1
0
∫
1
0
Si
B
(As − G a)d yd z =
1
3
∫
Ge
Si
dz
∫
1
0
∫
1
0
Ge
B
(P − Al ) d ydz =
2
3
∫
Ge
Si
dz
y
1
=
1
3
B
SiGa
ln(z)
y
2
=
2
3
B
SiGe
ln(z)
∇ ×
u = 14
i
∇ ×
v = 27
i
of 25 31
of 26 31
of 27 31
Appendix 1 (Atomic Radii, I used data set 4 for my calculations)
of 28 31
(The Densities, gram per cubic cm)
of 29 31
Appendix 2 (The Path Integral And Its Vector Form of Path)
of 30 31
of 31 31
The Author